Question

A 50.0 mL solution of 0.110 M KOH is titrated with 0.220 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl. 0.00 mL pH = 7.00 mL pH = | 12.5 mL pH = 20.0 mL pH = 24.0 mL pH = 25.0 mL pH = 26.0 mL pH = 29.0 mL pH =

Answer 1

The answer is given in the attachment.
Answer -4. O.llom kot & 50 ml. 0.110 x mewole cul. initially katt present: 55 mmole koll. 2 t 0.00 ml. so PH [OH-] = 0.22om Hel added. -. 505" - 0.41 (MJ KOT pots - log [OH-] = 0.981 ... pH & Poh = 14 or pH = 14-0.95 - 13.05. when 2. When 7 . nul of. 0.22m Hel added 7 x 0.22 mmole. = 1.54 mmole Hel. added KOH + Hel → H₂0 tkd initial 5.5 o o o 1054 1.54 (mmole = 3.96 Total volm : (7 +50) = 57 ut. Total immole = 3.96 tu= 0.07 m) or pot = 1.15. 5707 pH = 14- 115 - 12.85 (mmole) at eq-m 5.5-1.54 .: [OH-] = 3.96...

similarly When 12*5 = 2.75 mmole tid. ml 0.22 m Hel added added so kot refta 5.5-2.75 = 2.75 mmole Total volm of solm =(50+ 12.5) we = 62. ...[04.1s 217515 (wy. - 01044.[ng poh . . . 4:35. 62.5 4. 4,4 mmole tel Plt - 24-1035 = 12.65 When 20.0 ne .0.22 m. Hel added = Koh eeft. 5.5-4.4 = il nomola total nohm = (50 + 20 = 70 mL. [OH-] = fata s 0.016 polt a 1.79 70 pH 12:21 24.0 ne '0.22 M Heladded = 24 x 0.22 = 5.28 nmole kot left=(5.5-5.28) =(0.22) mumole Total volm=160 +24) = 74 mL TOHI = 0.22 74 297 x 103 m. poh 2:55: pH = 14-2.55 = 11.45

when 25.0 h 0.22 m Hel. added : 5.5 mmole tel. Kott Reft. (6.5 - 5.5) 30 This is equivalence point. at eqre point of strong acid and strong base pH ist. i I. when 26.0 ml 0.22 M Hel added a 5.72 mmol . Hel KOH initial a 5.5 nmole. KOH & Hel » Hetkel. so Hel excess in the solu = (5.72-5.5) (0.222 mmole. Total rohn of sohn = (26 +50) = T6 aut. [H +): 0:22 289 x 10-3(1) pH = -log H+ = 2.54 29 mi 0.22 M Hel added = 6.38 nmole Hel cucess a Coro-5.5) nemole a 0.88 mmole tolal volm = 2q + 50 = 79 ne. [HE] = tem = 0.01 (m) pH = 196 HC 8.